What is the reason for the guess?

Y-Y connection impedance are series.

Your vids are helpful but please get a new mic, your volume is so low.

FE exam has 110 questions, if you can save time on 1 question do it, if P=Q then angle=45 degree, end of the problem, on you calculator make a conversion from Pol (10, 45) to Rec and you have the answer immediately.

so what is the clock for?🤔

Thanks for the information, but the audio is too low, I can't hear anything even with max volume.

This answer is Wong I1= .6 , i2 = 1.1 , i3=.5

This answer is wrong I1 = .6 , i2=1.1 , i3 = .5 , power 1.1^2 = 1.21 watt

How can I get this FE Handbook?

What was you hiding from the mob when you recorded this? Thanks for vid though

The answer is correct. Another way to do it and make it more understandable : Line Voltage (VL) Phase Voltage (Vp) Line Current (IL) Phase Current (Ip) Z1= 7.07+j7.07 Z2= 7.07-j7.07 I1= Current Wye 1 I2= Current WyE 2 Io= Current from the source *Remembering Wye VL= √3*Vp IL= Ip First, we gotta find Vp ( We use phase voltage since IL=Ip) Vp= VL/√3 Vp= 99.88V For the Wye1 (Y1): I1= Vp/Z1 = 9.89∡-45 For the Wye (Y2) I2= Vp/Z2 = 9.89∡45 Since Y1 and Y2 are in parallel we can say with confidence that the current from the source is I1+I2 (See it as a node): Io=I1+I2= 14.127A

Why can you treat the wye resistors like they are one resistor each and just apply the parallel equation to them like a normal circuit? I'm trying to look up a law or rule than allows for that

You solve the problem as though it's a ring counter, and as though the diagram had the output tied to the input. According to your diagram and the lack of an input we would see zeros shifted in.

Vout would be -4 right?

can't see them asking this on the FE lol

Thanks 😊

why the voice tune is so low? hardly hear what is saying?

Where does the equation I(in) = V(in) / Z (in) come from? I cant seem to find it in the FE handbook, was wonder if it was derived from something?

Thanks for doing this

Very well explained. Thanks for putting these videos out

P=1222W

How come we're able to leave voltage in terms of its RMS value? Don't we usually express the magnitude of a signal in terms of its peak?

100/4=17?

Man, sorry to say that, but don’t tell me that you passed FE solving three phase system this way.

For Ideal transformer P in is equal to P out, so just calculating P1 will save you a lot of time. Plus I don’t think the final answer is 25w.

L2 is shared by both mesh, It should be l2-l1te the resistance

Excellent video, thank you

Im sorry to say but he's correct, you are all wrong in the comments. I understand why you're confused but he's actually doing it correctly. It would be more clear if you draw it out. This problem is much like when I was studying wye-start delta-run motor starters. Delta connected loads see one-third the power as wye connected loads for balanced systems. Solve the wye system and multiply by 3 to check if you're delta answer is correct.

I believe the angle should be negative to indicate a capacitive load. Other than that, great videos. Thank you for making them

I'm surprised these videos don't have more views. Very helpful, thanks

When it comes to semiconductors, you can call me rusty trombone

Good thing we have only 3 minutes to solve this problem.

how did you get Zt=2<45?

Why don’t you use the node method

A bit lost as to Zt=…before you get 2<45•

Is/Ip = Np/Ns, you assumed the secondary current was the same as the primary current

How did he get 2 <45°?

1/N where N is 3 not 20, N is the number of data points there are

your way is fine but i prefer to calculate the current of the line by doing 240/50 = 4.8A through the circuit and then you can calc the VS = IR or VS = (4.8)(10+50+10j) or VS=4.8(60+10j) == VS=(288+48j) take the M of that and that gives you 291.97 as well :D

are you sure you can "overlap" when doing this?

Good video !!

Great series. Anyone studying for FE and needing a quick example, this guy is great.

R_2^2/(R_2+jR_2)=R_2/(1+JR_2) not *R_2/(1+j). You need to multiple the reciprocal (1-jR_2) the both the top and bottom of the fraction to be left with R_2/1=R_2, making the abs value equal to 10 not *5sqroot(2).

Yeah, I think ZL = Zs*. my understanding is that the conjugate corrects the power factor.

For An equation, if the interval from π-2π was not zero, would I have to multiply the ft value by cos(nt) as well?

How did you -45 when both the real and imaginary parts of the impedance are both positive. Shouldn’t the angle just be 45 degrees?

you can also solve by another equation: PF = cos(arctan(Q/P)) then use the P = S(PF)

Thank you so much for these videos, it is so helpful.

If it’s B or not B then wouldn’t it always be 1 instead of cancelling them both?

What. Is. With. The. Sound.